Currently working on a django social media application where next to posting every post to the feed, the information of each upload should create a pdf document, containing caption, image, created_at and image_id.
I´ve put the canvas into the upload functions, so that both will be created on the same click. So far, i get a pdf (can't get the attributes from the post into the pdf tho) and the post is uploaded just fine.
How do I get the posted data into the pdf?
And how do save that pdf to a folder within the project instead of starting an automatic download? The user should not be able to notice the pdf converting. It is just for back office - but very necessary due to the social media website being a part of a big installation. So how do I get these pdfs?
Here is the code:
views.py
def upload(request):
if request.method == 'POST':
#user = request.user.username
image = request.FILES.get('image_upload')
caption = request.POST['caption']
new_post = Post.objects.create( image=image, caption=caption)
new_post.save()
#create pdf
buffer = io.BytesIO()
p = canvas.Canvas(buffer)
p.drawString(100, 100, "Hello world.")
p = request.FILES.get('post_pdf')
p.drawText('caption')
p.drawImage('image')
p.showPage()
p.save()
buffer.seek(0)
return FileResponse(buffer, as_attachment=True, filename='hello.pdf')
return redirect('/')
else:
return redirect('/')
models.py
class Post(models.Model):
id = models.UUIDField(primary_key=True, default=uuid.uuid4)
image = models.ImageField(upload_to='post_images')
caption = models.TextField(max_length=300)
created_at = models.DateTimeField(auto_now_add=True)
number_of_likes = models.IntegerField(default=0)
number_of_dislikes = models.IntegerField(default=0)
#interaction_count = models.IntegerField(default=0)
#engagement_count = models.IntegerField(null=True)#number_of_dislikes + number_of_likes
def __str__(self):
return self.caption
To add the value of Post.caption into the pdf, use the value of new_post.caption, change this:
p.drawText('caption')
for this:
p.drawText(new_post.caption)
Same for other fields.
This is not as hard as it seems,
so Let's see if you are successful in creating a pdf and now you have to store it in background instead of downloading.
file_name = request.FILES["file_name"]. #suppose file_name is file name
file_name = default_storage.save(rf"{basePath}/media/whatsapp/file_name.pdf", file_name) #{basePath}/media/whatsapp/ is the path name where we want it to be stored
Related
This code return a TypeError as expected str, bytes or os.PathLike object, not InMemoryUploadedFile
I don't know how to pass user data in the form of file and image to my code.py file for making changes to the original.
views.py
def home(request):
new_image = None
file = None
form = ScanForm()
if request.method == 'POST':
form = ScanForm(request.POST, request.FILES)
if form.is_valid():
image = request.FILES['image']
xml_file = request.FILES['xml_file']
new_image = code.create(image, code.search(
xml_file)[0], code.search(xml_file)[1])
form.save()
return render(request, 'app/home.html', {'form': form, 'new_image': new_image})
else:
form = ScanForm()
return render(request, 'app/home.html', {'form': form, 'new_image': new_image})
printing image and xml_file successfully prints out their names
forms.py
class ScanForm(forms.ModelForm):
class Meta:
model = Scan
fields = '__all__'
models.py
class Scan(models.Model):
image = models.ImageField(upload_to='images')
xml_file = models.FileField(upload_to='files')
processed_at = models.DateTimeField(auto_now_add=True)
description = models.CharField(max_length=500, null=True)
class Meta:
ordering = ['-processed_at']
def __str__(self):
return self.description
Here is the code for manipulation of image according to the data in the xml
code.py
def search(path):
new = []
object_names = []
object_values = []
txt = Path(path).read_text()
txt.strip()
names = et.fromstring(txt).findall('object')
for i in names:
object_names.append(i[0].text)
values = et.fromstring(txt).findall('object/bndbox')
for i in values:
for j in i:
object_values.append(int(j.text))
return object_names, object_values
def create(image, object_names, object_values):
img = cv.imread(image)
on = len(object_names)
ov = len(object_values)
for i in list(range(0, ov, on)):
cv.rectangle(img, (object_values[i], object_values[i+1]),
(object_values[i+2], object_values[i+3]), (0, 0, 255), thickness=5)
return img
This code.py works fine if tested by passing data manually using local folder.
Here is the Traceback:
Traceback image
pathlib.Path() handles file paths, not memory objects. request.FILES are the data attached to the POST request. During your handling of a POST request, you can validate this data and decide to save it to the server disk.
If you would like your image processing to read the file from the server disk, you have to save the new model instance first. You can then access the file's path on disk through the name attribute of the model's ImageField, see Using files in models.
If you want to handle the uploaded data before saving it to disk, you can read it as follows:
txt = request.FILES["xml_file"].read()
See UploadedFile.read()
I am working on a platform where I need to allow users to upload multiple images in one post.
I want to keep it as simple as possible, so that a person wouldn't have to refresh the page to upload each image, or create and save a post before adding images.
If a user could delete or reorder images it would be nice
Can you give me advice on how to do it properly?
I am using django , postgres and here's what I have done so far .
On my models.py -
class ImagePost(models.Model):
user = models.ForeignKey("profiles.HNUsers", on_delete=models.DO_NOTHING)
image = OptimizedImageField("Post Image", blank=True, null=True)
timestamp = models.DateTimeField("Timestamp", blank=True, null=True, auto_now_add=True)
text = models.TextField("Description text", blank=True)
class Meta:
verbose_name_plural = "Image Posts"
It can take one image just fine . what should I change to make it multi upload post
On my views.py this is what I have done so far -
#api_view(['POST'])
#permission_classes((permissions.AllowAny,))
def image_post(request):
if request.method == 'POST':
data = request.data
print(data)
serializer = ImagePostSerializer(data=data)
if serializer.is_valid():
serializer.save()
print("image object saved")
try:
image_post_object = ImagePost.objects.filter(user__id=data['user']).order_by('-timestamp')[0]
print(image_post_object)
try:
post = Posts()
post.user = HNUsers.objects.get(id=data['user'])
post.image_url = image_post_object.image.url
post.type = 'I'
post.category = data['category']
post.created_on = image_post_object.timestamp
post.text = image_post_object.text
save_post = post.save()
post_id = post.pk
try:
user = HNUsers.objects.get(pk=data['user'])
if user.user_type == 'HN':
payload = {
"user_name": user.full_name,
"image_url": user.profile_img_url,
"post_id": post_id,
"notification_type": "hn"
}
print(payload)
broadcast_notification('all', payload, 1)
except HNUsers.DoesNotExist:
print("user_id_does_not_exist")
print("post object created")
except Posts.DoesNotExist:
print("Failed - post creation failed")
except HNUsers.DoesNotExist:
print("Failed - user object not found")
except ImagePost.DoesNotExist:
print("Failed - image object not found")
return Response(serializer.data, status=status.HTTP_200_OK)
Any advice is really appreciated.
I am not sure if this can help, but why you dont use more than one table to store images in DB:
class Images(models.Model):
user = models.ForeignKey("profiles.HNUsers",
on_delete=models.DO_NOTHING)
image_post = models.ForeignKey(ImagePost, on_delete=models.CASCADE)
image = OptimizedImageField("Post Image", blank=True, null=True)
Now you have conected multiples images to a post and a user.
Hope this helps a bit...
Check this excellent django version of jQuery-File-Upload :
https://github.com/sigurdga/django-jquery-file-upload
Some features:
Drag and drop files
Select multiple files
Cancel upload
Delete uploaded file
You can avoid a lot of code repetition and add multiple images for a model by creating an intermediate model which will handle mapping between your image and the model. Create proper binding like one to one , you can look at this tutorial
models.py
class FileUpload(models.Model):
File_Name = models.CharField(max_length=255, blank=True)
File_path = models.FileField(upload_to='')
Description = models.CharField(max_length=255, blank=True)
Upload_Date = models.DateTimeField(auto_now_add=True)
forms.py
class FileUploadForm(forms.Form):
class Meta:
model = FileUpload
File_Name = forms.CharField(label="File Name",max_length=255)
Description = forms.CharField(label="Description", max_length=255)
I'm new in Django.I need help. How to upload images in the database and view those images? Thanks in advance!
here paths are stored in database and images are stored in a folder. But I don't need that. I want to save images and path to the database and I need to view that image. Please help!
views.py:
def uploadfile(request):
print('inside upload logic')
if request.method == 'POST':
form = FileUploadForm(request.POST, request.FILES)
if form.is_valid():
# ImageUpload(request.FILES['File_Name'])
myfile = request.FILES['File_Name']
fs = FileSystemStorage()
filename = fs.save(myfile.name, myfile)
uploaded_file_url = fs.url(filename)
newdoc = FileUpload(File_Name=myfile.name, File_path=uploaded_file_url, Description=request.POST['Description'])
newdoc.save()
#return HttpResponse("File uploaded successfuly")
return render(request, 'Login/fileupload.html')
else:
form = FileUploadForm()
return render(request, 'Login/fileupload.html', {
'form': form
})
You normally shouldn't store the image data in your database. If you need to upload and store images, use the ImageField or FileField and follow the instructions from the docs to save the image. The only thing you need to do is:
form = FileUploadForm(request.POST, request.FILES)
if form.is_valid():
uploaded_file = FileUpload(
File_path=request.FILES['File_path'], # or whatever you've called the file input
File_name=form.cleaned_data['File_Name'],
Description=form.cleaned_data['Description'])
uploaded_file.save()
It would be easier to use a ModelForm in your case, then you only need to save the form:
if form.is_valid():
form.save()
This will automatically save the file in the correct way. No need to do the saving manually like you are doing.
To view the image is as simple as this:
file_upload = FileUpload.objects.get(id=34)
file_url = file_upload.File_path.url # url relative to `MEDIA_ROOT`
# or in a template
{% load static %}
{% get_media_prefix %}{{ file_upload.File_path.url }}
If you really need to store the images as binary blobs to your database (but beware that it almost never makes sense to do so), use Django's BinaryField, as described here.
This also means you will have to handle transforming the image back and forth from a binary blob to a proper image file. Consider also storing the content type (image/jpg or image/png or image/webp) since you will need that to properly re-create the file.
I have the following model.
class Image(models.Model):
customer = models.ForeignKey(Customer, related_name='images')
image = models.ImageField(upload_to='/pictures/', verbose_name='Image')
Each time user add a new Image I want it to go under pictures/customer.id/custom_filename
When I was using local filesystem, that was easy. I used a function that handled file chunks upload with a new name and returned the new path. But now I want to use S3 to store my files. So I am using django-storages. I did the following tests:
testmodel
class TestModel(models.Model):
name = models.CharField(max_length=10)
logo = models.ImageField(upload_to='pictures/')
when I do this in my view
def index(request):
form = TestModelForm(request.POST or None, request.FILES or None)
if request.method == 'POST':
if form.is_valid():
print 'posting form'
model = form.save(commit=False)
print model.name
print model.logo.url
model.save()
and the image is uploaded as it should under mybucket.s3.amazon.com/pictures/logoname.jpg
But if I change the file name by some way like
def index(request):
form = TestModelForm(request.POST or None, request.FILES or None)
if request.method == 'POST':
if form.is_valid():
print 'posting form'
model = form.save(commit=False)
print model.name
filename = str(request.FILES['logo']).split('.')[0]
extension = str(request.FILES['logo']).split('.')[1]
path = '%d%d%d_%s_%d.%s' %(datetime.now().year, datetime.now().month, datetime.now().day, filename, models_no+1, extension)
model.logo = path
print model.logo.url
model.save()
i get a new url which is mybucket.s3.amazon.com/newlogoname.jpg (correct since I didn't user pictures in the new path) but the file is not uploaded. Must I do it manually using boto? I guess it's the only way since folders for each customer (initial example) might not exist and need to be created. What is the correct way to upload images at custom locations/directories for each model?
I tried to call default storage directly to save the image in the location I wanted but didn't work
default_storage.save('pictures/one.jpg', ContentFile(request.FILES['logo']))
The key is created (the folder and the file) but the image is empty. I think using cloud storage in a flexible way is very difficult :(
You can pass a callable as the upload_to named argument, in your model definition (so you actually don't have to do this on the view level). To make it clear, try doing something like this:
class Image(models.Model):
customer = models.ForeignKey(Customer, related_name='images')
image = models.ImageField(upload_to=self.generate_upload_path, verbose_name='Image')
def generate_upload_path(self, filename):
return os.path.join('/pictures/'+str(self.customer.id), filename)
According to the official django documentation about uploads, small files are saved to memory and big files are saved to disk.
I would like to know how to save uploaded files to disk? Is it possible?
This is my code so far. But it only works on memory. When I try to write the file to disk, or the file is big in size the app crashes.
views.py
# ...
def spreadsheet_form(request, id = None):
if is_admin_user(request):
instance = get_object_or_404(Spreadsheet, id=id) if id is not None else None
form = SpreadsheetForm(request.POST or None, request.FILES or None, instance=instance)
if form.is_valid():
spreadsheet = form.save(commit=False)
spreadsheet.name = request.POST['name']
spreadsheet.spreadsheet_file = request.FILES['spreadsheet_file'].name
spreadsheet.size = request.FILES['spreadsheet_file'].size
spreadsheet.save()
handle_uploaded_file(request.FILES['spreadsheet_file'])
return redirect('/spreadsheets/')
return render_to_response("pages/spreadsheet_form.html", {"form": form,"id":id},context_instance=RequestContext(request))
else:
return redirect('/', False)
# ...
def handle_uploaded_file(f):
with open(f.name, 'wb+') as destination:
for chunk in f.chunks():
destination.write(chunk)
models.py
# ...
class Spreadsheet(models.Model):
id = models.AutoField(primary_key=True)
name = models.CharField(max_length=256)
spreadsheet_file = models.FileField(upload_to='spreadsheets/')
size = models.CharField(max_length=32)
created_at = models.DateTimeField(auto_now=True)
# ...
def __unicode__(self):
return u'%s' % (self.name )
settings.py
FILE_UPLOAD_HANDLERS = (
"django.core.files.uploadhandler.MemoryFileUploadHandler",
"django.core.files.uploadhandler.TemporaryFileUploadHandler",
)
FILE_UPLOAD_MAX_MEMORY_SIZE = 7000000
FILE_UPLOAD_TEMP_DIR = '/tmp'
Thanks in advance
There's no such thing as 'save to disk' in the App Engine world. The closest was Blobstore, and now it's GCS.
django-nonrel includes a django storage class to upload to Blobstore. Follow this:
http://www.allbuttonspressed.com/blog/django/2010/06/Uploads-to-Blobstore-and-GridFS-with-Django
You don't need to fiddle with the FILE_UPLOAD_HANDLERS, the defaults are fine.